Laws characterizing the trivial group
Equivalences:
the law $w$ characterizes the trivial group among all groups;
the law $w$ characterizes the trivial group among all cyclic groups of prime order;
the image of $w$ in the abelianization of the free group is a primitive element.
That 1 implies 2 is trivial. Suppose conversely that 1 fails. Then there exists a nontrivial group $G$ satisfying the law $w$. Being nontrivial, $G$ has a nontrivial cyclic subgroup, and the latter has a quotient of prime order. Since satisfying a law passes to subgroups and quotients, we deduce that this group of prime order satisfies the law $w$, showing the failure of 2.
If 3 fails, then $w\in [F,F]F^p$ for some prime $p$ and hence $w$ holds in the cyclic group of order $p$, showing failure of 2. Conversely, assume 3. A primitive element of a free abelian group can be mapped by an automorphism to the first basis vector, and hence by an homomorphism to the generator of the infinite cyclic subgroup, and hence after composition to any element of any group. Let now $G$ be any nontrivial group; then we have an homomorphism $F\to G$ mapping $w$ to a nontrivial element, so $G$ fails to satisfy $w$. Thus $w$ characterizes the trivial group among all groups.
Every word $w$ on free generators $x_1,\dots,x_n$ can be written as $$w=x_1^{\alpha_1} \cdots x_n^{\alpha_n} c(x_1,\dots,x_n),$$ where $c$ is a word in the commutator subgroup of $\langle x_1,\dots,x_n \rangle$ and $\alpha_1,\dots, \alpha_n$ are some integers. Note that the integer $\alpha_i$ is the same as the sum of the exponents of letter $x_i$ in the word $w$. You have denoted $\alpha_i$ by $\deg_{x_i}(w)$.
Now we can state our characterization of such words:
A word $w$ is charaterizing the trivial group if and only if $\gcd (\deg_{x_1}(w),\dots,\deg_{x_n}(w) )=1$.
This is because if $d:=\gcd(\alpha_1,\dots,\alpha_n)\neq 1$, then the cyclic group of order $d\neq 1$ satisfies $w$. Now consider $w(1,\dots,a,\dots,1)$, where $a$ is an arbitrary element of the group satisfying the law $w$ and $a$ is in the possition $i$. It follows that $a^{\alpha_i}=1$ for all $i=1,\dots,n$ (note that $c(1,\dots,a,\dots,1)=1$). Thus $a^d=1$ and so $a=1$.